How to show that the mild solution to this parabolic equation is also a classical solution? Let $U\subseteq \mathbb{R}^n$ be a smooth bounded domain and consider the following problem:
$$
\begin{cases}
\partial_t u = \Delta u + u &\text{in } U\times (0, \infty)\\
u = 0 &\text{on } \partial U \times (0, \infty)\\
u = g &\text{on } U\times \{0\},
\end{cases}
$$
where $g \in C(\bar{U})$ is given.
I want to prove that this system admits a solution $u \in C^2(U\times (0, \infty))\cap C(\bar{U}\times [0, \infty))$. 
To do this, I constructed an operator $A$ mapping from a subset of $H_0^1(U)$ to $L^2(U)$. More precisely, I defined $A$ so that
$$Au = f \iff
\int_U \nabla u \cdot \nabla v - uv = \int_U fv \qquad\forall v\in H_0^1(U).
$$
So far, I have  been able to show that this operator $A$ has eigenfunctions $(u_k)_k$ forming an orthonormal basis of $L^2(\Omega)$. Furthermore, the corresponding eigenvalues $(\lambda_k)_k$ are bounded from below and $\lambda_k \to \infty$ as $k\to\infty$.
Next, let us define the function
$$
u(x,t) = \sum_{k=1}^\infty e^{-\lambda_k t}\langle u_k, g\rangle_{L^2(U)} u_k(x).
$$
One can show that the function $v:[0, \infty) \to L^2(\Omega)$ given by $v(t) = u(\cdot, t)$ satisfies the following properties


*

*$v\in C\left([0, \infty), L^2(\Omega)\right)$,

*$v \in C^1\left((0, \infty), L^2(\Omega)\right)$,

*$v(0) = g$ in the $L^2$-sense,

*$v^\prime(t) = -Av(t)$ for all $t>0$.


I am sure that the method I am using has a name. If someone could either elaborate on it here or point me to a source where I could read further it would be highly appreciated. As of here, I have no idea on how to proceed. Any comments are welcome.
 A: What you describe is a semigroup approach to evolution equations. In general, if $H$ is a Hilbert space and $A$ a lower bounded self-adjoint operator on $H$, then for every $g\in H$ there exists a unique mild solution to the initial value problem
\begin{align*}
\begin{cases}\partial_t v(t)=-Av(t)\\
v(0)=g\end{cases}
\end{align*}
given by $v(t)=e^{-tA} g$. Moreover, $u$ is a classical solution in the sense that $v\in C^1((0,\infty);H)$ (in fact, $v\in C^\infty((0,\infty);D(A^k))$ for all $k\in\mathbb{N}$). In the case when $A$ has pure point spectrum, then the operator exponential $e^{-tA}$ can be expressed in terms of the eigenfunctions in the way you describe.
As you see, one gets regularity in the time variable "for free" from abstract theory. However, one cannot expect to get any regularity in the space variable in this abstract setting (assuming $H=L^2(\Omega))$. For example if $Ag=0$, then $v(t)=g$ is a solution of the initial value problem, but the eigenfunctions of a self-adjoint operator on $L^2$ do not need to have any additional regularity.
The situation is much better if $A$ is for example (a self-adjoint realization of) a uniformly elliptic operator of order $m$ with smooth coefficients. Elliptic regularity theory implies that $D(A^k)\hookrightarrow H^{km}(\Omega)$. Since $v\in C^\infty((0,\infty);D(A^k))$ for all $k\in\mathbb{N}$, we have $v\in C^\infty((0,\infty);H^l(\Omega))$ for all $l\in\mathbb{N}$. The Sobolev embedding theorem yields $v\in C^\infty((0,\infty);C^l(\overline\Omega))$ for all $l>0$. Now you can define $u(t,x)=v(t)(x)$ and convince yourself that $u\in C^\infty((0,\infty)\times \Omega)$ and that the initial value problem is satisfied in the classical sense.
I think a good starting point for beginners is Brezis's Functional Analysis, Sobolev Spaces and Partial Differential Equations. The argument I gave is very similar to the discussion of the heat equation in Chapter 10 of this book.
